In the previous section, we saw the basics of linear inequalities in two variables. We saw a solved example also. In this section, we will see a few more solved examples.
Solved example 6.10
Solve 3x+2y>6 graphically
Solution:
1. We want to solve the inequality 3x+2y > 6
• Let us first compare it with the standard form: ax +by < c
♦ We see that, b is +ve. And the sign before c is '>'
♦ So based on table 6.1, we get: Upper plane II
• The table 6.1 is shown again below for easy reference:
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| Table 6.1 |
2. In fig.6.12 below, the red dashed line is the graph of 3x + 2y = 6
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| Fig.6.12 |
• Dashed line is used instead of bold line because, the sign in the given inequality is is '>' not '≥'
• This line divides the Cartesian plane into Lower half plane I and upper half plane II
• Based on the result in (1), all points in the upper plane II will satisfy the inequality 3x+2y > 6
3. We can check this using any convenient point. The point O(0,0) is convenient for calculations. So we will use that point.
• If we input (0,0) in the inequality, we will get:
♦ (3 × 0) + (2 × 0) > 6
♦ ⇒ 0 + 0 > 6
♦ ⇒ 0 > 6, which is not true.
4. O(0,0) lies in the lower half plane I. So the check in (3) confirms that, none of the points in the lower half plane I will satisfy the inequality 3x+2y > 6
• So it is clear that the other half plane II is the required area. We hatch that area using green lines. The result that we obtained in (2), using table 6.1 is correct.
5. We see that, the upper half plane II is hatched. All points in that half plane will satisfy the given inequality.
• But any half plane will be an infinite area. We cannot hatch it completely on a sheet of paper or on a computer screen.
• So we write:
Graph of 3x+2y > 6 is the upper half plane. Points on the line 3x+2y=6 are not included.
Solved example 6.11
Solve 3x-6 ≥ 0 graphically
Solution:
1. We want to solve the inequality 3x-6 ≥ 0
• Let us first write it in the standard form: ax +by ≥ c
We get: 3x + 0y ≥ 6
♦ We see that, b is neither +ve nor -ve.
♦ So we cannot use table 6.1
♦ We must check using a convenient point.
2. In fig.6.13 below, the red line is the graph of 3x = 6
 |
| Fig.6.13 |
• This line divides the Cartesian plane into left half plane I and right half plane II
3. We can find the appropriate plane by using any convenient point. The point O(0,0) is convenient for calculations. So we will use that point.
• If we input (0,0) in the inequality, we will get:
♦ (3 × 0) ≥ 6
♦ ⇒ 0 ≥ 6, which is not true.
4.
O(0,0) lies in the left half plane I. So the check in (3) confirms
that, none of the points in the left half plane I will satisfy the
inequality 3x ≥ 6
• So it is clear that the other half plane
II is the required area. We hatch that area using green lines.
5. We see that, the right half plane II is hatched. All points in that half plane will satisfy the given inequality.
• But any half plane will be an infinite area. We cannot hatch it completely on a sheet of paper or on a computer screen.
• So we write:
Graph of 3x-6 ≥ 0 is the right half plane. Points on the line 3x=6 are also included.
Solved example 6.12
Solve y < 2 graphically
Solution:
1. We want to solve the inequality y < 2
• Let us first write it in the standard form: ax +by < c
We get: 0x + y < 2
♦ We see that, b is +ve. And the sign before c is '<'
♦ So based on table 6.1, we get: Lower plane I
2. In fig.6.14 below, the red dashed line is the graph of y = 2
 |
| Fig.6.14 |
• Dashed line is used instead of bold line because, the sign in the given inequality is '<' not '≤'
• This line divides the Cartesian plane into Lower half plane I and upper half plane II
• Based on the result in (1), all points in the lower half plane I will satisfy the inequality y < 2
3. We can check this using any convenient point. The point O(0,0) is convenient for calculations. So we will use that point.
• If we input (0,0) in the inequality, we will get:
♦ 0 < 2, which is true.
4.
O(0,0) lies in the lower half plane I. So the check in (3) confirms
that, all points in the lower half plane I will satisfy the
inequality y < 2. We hatch that area using green lines. The
result that we obtained in (2), using table 6.1 is correct.
5. We see that, the lower half plane I is hatched. All points in that half plane will satisfy the given inequality.
• But any half plane will be an infinite area. We cannot hatch it completely on a sheet of paper or on a computer screen.
• So we write:
Graph of y < 2 is the lower half plane I. Points on the line y = 2 are not included.
The link below gives some more solved examples:
Exercise 6.2• In the next section, we will see system of linear inequalities.
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